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Dmitriy789 [7]
3 years ago
9

Where are the minimum and maximum values for f(x) = -2 + 4 cos x on the interval (0,21]?

Mathematics
2 answers:
Dafna11 [192]3 years ago
4 0

Answer:

Step-by-step explanation:

Maximum value is when cos x = 1

So it is -2 + 4(1) = 2.

Minimum value, when  cos x = -1:

= -2  + 4(-1) = -6.

Alex_Xolod [135]3 years ago
4 0

Answer:

The maximum 2 is reached when x=2pi,4pi, and 6pi.

The minimum -6 is reached when x=pi, 3pi,and 5pi.

Step-by-step explanation:

So let's first look at cos(x) on interval (0,21].

How many rotations is that? Does it at least contain 1 full rotation? If it contains one full rotation that means all the values from -1 to 1 (inclusive) are tagged? If it doesn't contain a full rotation, we might have to dig a little deeper.

So we know x=0 isn't included and that's when cosine is first 1,but this doesn't mean 1 won't be hit later.

Let's figure out the number of rotations:

21/(2pi)=3.3 approximately

This means we make at least 3 rotations.

So this means we definitely will have all the values from -1 to 1 tagged (inclusive).

Now let's look at whole function.

f(x) = -2 + 4 cos x

-2+(-4) to -2+4 will be the range of the function

So the minimum is -6 and the maximum is 2.

So the min occurs when cos(x)=-1 and the max occurs when cos(x)=1.

We have a little over three rotations and remember we can't include x=0.

cos(x)=1

when x=2pi (one full rotation)

when x=4pi (two full rotations)

when x=6pi (three full rotations)

We will stop here because cosine won't be 1 again until a fourth full rotation

cos(x)=-1

when x=pi (half rotation)

When x=3pi (one + half rotation)

When x=5pi (two+half rotation)

We can't include x=7pi (three+half rotation)

because this one is actually not in the interval because 3.5 is more than 3.3 .

The maximum 2 is reached when x=2pi,4pi, and 6pi.

The minimum -6 is reached when x=pi, 3pi,and 5pi.

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3 years ago
Which statement best reflects the solution(s) of the equation?
Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1} and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0

Now, factoring the term:

x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}  the denominator becomes zero i.e

\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2  in the equation,

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}

\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

  • brainly.com/question/1626495
  • brainly.com/question/2959656
  • brainly.com/question/2456302

#learnwithBrainly

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-8w+3z
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The answer is B. It is shifted up by 6*1.5-2-2 with is 9-2-2=5.

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